i seriously don't know how to do these..and i really need help..
$\displaystyle \frac{tanx sinx}{tanx-sinx}=\frac{tanx+sinx}{tan sinx}$
$\displaystyle sin^3\theta (1+cot\theta) + cos^3\theta (1+tan\theta)=sin\theta+cos\theta$
For the first one, no one side is more complicated than the other, so it probably doesn't matter which side you start with. Let's say we start with the left-hand side. Try converting everything to sines and cosines, and see where that leads.
For the second one, a good first step will be to multiply things out on the left-hand side, and see where that leads you....
On the first I am going to work left to right. We can start by writing the left side as:
$\displaystyle \frac{tanx sinx}{tanx-sinx}
$
As
$\displaystyle \frac{\frac{sinx sinx}{cosx}}{\frac{sinx}{cosx}-sinx}$
From here I can rearrange the fraction as:
$\displaystyle \frac{sinx sinx}{cosx}\frac{cosx}{sinx - cosxsinx}$
Which is really:
$\displaystyle \frac{(sinx)^{2}}{sinx - cosxsinx}$
And on the bottom I can factor:
$\displaystyle \frac{(sinx)^{2}}{sinx(1 - cosx)}$
See where you go from here. . .
On the second one, the left side is a lot easier to work with than the right side (I mean where can you go from that?). I would start by multiplying out, and then doing some cancellations. You can also work with the right side to see if it allows you to meet the left side somewhere in the middle (but do this as sidework, not showwork.)
Cancel off the common factor. Then look at the denominator.
Often (though of course not always) it can be helpful to multiply that sort of denominator by the conjugate. In this case, multiply the fraction, top and bottom, by 1 + cos(x). Simplify.
Do not factor out and cancel the sines! Instead, try dividing through, top and bottom, by cos(x). Simplify, and see what you get.